访问嵌套数据 PHP - 编程之家

如何解决访问嵌套数据 PHP

我确定我一定遗漏了一些明显的东西，但我正在尝试从以下数据中访问一些场地和类别信息并使用 PHP 检索它。

当我在 foreach 语句中的 ['response'] 旁边包含 ['venues'] 时，我可以轻松访问名称、纬度和经度，没有问题。但是，我还需要从 ['categories'] 访问一些信息，所以我尝试将 ['venues'] 从 foreach 语句移到访问数组该部分的每一行上。这会导致一个错误，即 ['venues'] 和 ['categories'] 的每个实例都是一个未识别索引

我错过了什么？

我正在使用的数据：

data: {
    Meta: {
        code: 200,requestId: "61044a3db649b50f959a1dd9"
    },response: {
        venues: [
            {
                id: "4ac518cdf964a520e6a520e3",name: "National gallery",location: {
                    address: "Trafalgar Sq",lat: 51.50887601013219,lng: -0.1284778118133545,labeledLatLngs: [
                        {
                            label: "display",lng: -0.1284778118133545
                        }
                    ],postalCode: "WC2N 5DN",cc: "GB",city: "London",state: "Greater London",country: "United Kingdom",formattedAddress: [
                        "Trafalgar Sq","London","Greater London","WC2N 5DN","United Kingdom"
                    ]
                },categories: [
                    {
                        id: "4bf58dd8d48988d18f941735",name: "Art Museum",pluralName: "Art Museums",shortName: "Art Museum",icon: {
                            prefix: "https://ss3.4sqi.net/img/categories_v2/arts_entertainment/museum_art_",suffix: ".png"
                        },primary: true
                    }
                ],etc

我的 PHP：

?PHP

$url = "https://api.foursquare.com/v2/venues/search?client_id=foo&client_secret=bar&v=20210703&near=" . $_REQUEST['city'] . "," . $_REQUEST['country'] . "&categoryId=" . $_REQUEST['category'];

$curl = curl_init();

$ch = curl_init();
    curl_setopt($ch,CURLOPT_SSL_VERIFYPEER,false);
    curl_setopt($ch,CURLOPT_RETURNTRANSFER,true);
    curl_setopt($ch,CURLOPT_URL,$url);
   
    $result=curl_exec($ch);

    curl_close($ch);

    $decode = json_decode($result,true);

    $countryData = [];

    foreach($decode['response'] as $result){
        $countryinfo = [];
        $countryinfo['name'] = $result['venues']['name'];
        $countryinfo['lat'] = $result['venues']['location']['lat'];
        $countryinfo['lng'] = $result['venues']['location']['lng'];
        $countryinfo['icon'] = $result['categories']['icon']['prefix'];
        $countryinfo['iconSuffix'] = $result['categories']['icon']['suffix'];

         array_push($countryData,$countryinfo);
    }

 
    $output['status']['code'] = "200";
    $output['status']['name'] = "ok";
    $output['status']['description'] = "success";
    $output['data'] = $countryData;
    
    header('Content-Type: application/json; charset=UTF-8');

    echo json_encode($output); 


?>

如果我在 foreach 语句中包含场地，我会得到所需的名称、纬度和 lng 结果 - 所以我不知道为什么将它取出并放在访问场地数组的每一行上不起作用？

PHP：

foreach($decode['response']['venues'] as $result){
        $countryinfo = [];
        $countryinfo['name'] = $result['name'];
        $countryinfo['lat'] = $result['location']['lat'];
        $countryinfo['lng'] = $result['location']['lng'];
        //$countryinfo['icon'] = $result['categories']['icon']['prefix'];
        //$countryinfo['iconSuffix'] = $result['categories']['icon']['suffix'];

         array_push($countryData,$countryinfo);
    }

{
status: {
code: "200",name: "ok",description: "success"
},data: [
{
name: "National gallery",lng: -0.1284778118133545
},{
name: "Elizabeth Tower (Big Ben) (Big Ben (Elizabeth Tower))",lat: 51.50070194764357,lng: -0.12458056211471556
},``` etc